91Ó°ÊÓ

When n=0, the binomial theorem simplifies to: $$(a+b)^0 = \sum_{k=0}^0 {0 \choose k} a^k b^{0-k}$$ According to the definition of exponents, any non-zero value raised to the power of zero is equal to 1. Thus, \((a+b)^0 = 1\). Also, the only term in the summation (k=0) is \({0 \choose 0} a^0 b^0\). Since the combination \({0 \choose 0}\) is equal to 1 and \(a^0 = b^0 = 1\), the term evaluates to 1. So the equality holds for the base case.

Now, we assume that the binomial theorem is true for n, i.e., $$(a+b)^n = \sum_{k=0}^n {n \choose k} a^k b^{n-k}$$ Our goal is to show that the theorem is also true for \(n+1\), i.e., $$(a+b)^{n+1} = \sum_{k=0}^{n+1} {n+1 \choose k} a^k b^{(n+1)-k}$$ Using the definition of exponentiation, $$(a+b)^{n+1} = (a+b)^n \cdot (a+b)$$ Now, we use our assumption and multiply each term in the sum by \((a+b)\): $$(a+b)^n \cdot (a+b) = \left(\sum_{k=0}^n {n \choose k} a^k b^{n-k}\right)(a+b)$$ Distribute \((a+b)\) to each term in the sum: $$= \sum_{k=0}^n {n \choose k} a^{k+1} b^{n-k} + \sum_{k=0}^n {n \choose k} a^k b^{(n-k)+1}$$ Now, we can reduce the exponents of 'b' in the second summation by changing the index: $$= \sum_{k=1}^{n+1} {n \choose k-1} a^k b^{(n+1)-k} + \sum_{k=0}^n {n \choose k} a^k b^{(n+1)-k}$$ We can combine these two summations into one: $$= \sum_{k=0}^{n+1} \left[{n \choose k-1} a^k b^{(n+1)-k} + {n \choose k} a^k b^{(n+1)-k}\right]$$ Now, we can apply the Pascal's identity \({n \choose k-1} + {n \choose k} = {n+1 \choose k}\): $$= \sum_{k=0}^{n+1} {n+1 \choose k} a^k b^{(n+1)-k}$$ Since we have shown that the theorem holds for \(n+1\), we have completed the induction. Therefore, the binomial theorem is true for all \(n \in \mathbb{N}_0\).